Math Is Fun Forum

Identity

Member

Registered: 2007-04-18

Posts: 934

Hard problems

1. Express the number 1/2006 as the sum of the reciprocals of distinct integers of size less than 2006 - that is,

,

where -2006 <

< 2006 for all i = 1,2, ... , n.

Well... all I really know is that the denominators must multiply to a multiple of 2006... so must be a factor of a multiple of 2006... but what multiple...?
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2. Find all positive integers less than 10,000 which are equal to five times the product of their digits.

Guess that means:

I've already worked out that d = 5. This is because 1) the number must be a multiple of 5, and 2) no digit may be zero.
Next, every digit must be odd. This is because if any even number is multiplied by 5, the last digit of the product will be 0, and this is not allowed.
So we have:

, where

Don't know how to go further...
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These are the final two 'puzzle' type questions I'm having trouble with, in addition to the ones I've already posted... preparation for a math test on Wednesday... thanks.

Last edited by Identity (2007-04-28 01:12:16)

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Hard problems

For 1, I'm not sure how you could proceed other than with trial and error. You're right that the denominators have to multiply to give a multiple of 2006, which means that the factors 2, 17 and 59 must appear in there somewhere, but I don't know where to go after that.

For 2, you rightly say that the number must be a multiple of 5, because it's 5 times bigger than the product of its digits. However, fixing the last digit at 5 means that the product of its digits is also a multiple of 5, which means that you can deduce something further about the number itself. After that it gets a bit trickier, but you've narrowed the options down quite a bit by then so if nothing else, just using brute force becomes an option.

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Why did the vector cross the road?
It wanted to be normal.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Hard problems

For #1, do the integers really have to be distinct? If not, you could try this.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Hard problems

But that makes the problem trivial! I would think that the integers do have to be distinct. This problem is kind of similar to finding Egyptian fractions, but slightly different because negatives are involved.

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Why did the vector cross the road?
It wanted to be normal.

Zhylliolom

Real Member

Registered: 2005-09-05

Posts: 412

Re: Hard problems

1. To make this simple, let us see if we can do the problem with only two reciprocals, so 1/2006 = 1/a + 1/b, with -2006 < a, b < 2006. Multiplying both sides by 2006ab gives ab = 2006a + 2006b. We manipulate the expression so that we can factor:

Since -2006 < a, b < 2006, we have that 0 < 2006 - a < 2 × 2006 and 0 < 2006 - b < 2 × 2006. So we want two factors (2006 - a and 2006 - b) of 2006² that are between 0 and 2 × 2006. We know that 2006 = 2 × 17 × 59. So our factors are 2² × 17² (= 1156) and 59² (= 3481). Then 2006 - a = 1156 and 2006 - b = 3481, so a = 850 and b = -1475. Thus

2. You're perfect so far . The next step is to realize that 5 times the digits of our number (I'll call it N) must be divisible by 25, since 5 is its last digit. The multiples of 25 end in 00, 25, 50, or 75. Since N can't have 0 or an even number as a digit, it must end in 75. Once again, since N is 5 times the product of its digits, it is divisible by 5 × 7 × 5 = 175. Starting at 1 × 175, every 4th multiple of 175 ends in 75. So 1 × 175 = 175, 5 × 175 = 875, 9 × 175 = 1575, ..., 57 × 175 = 9975 (we stop at 57 since N must be less than 10000). If you write out all these products you see that the only numbers there with only odd digits are 175, 1575, 5775, 7175 and 9975. By direct calculation the only one of these which is equal to 5 times the product of its digits is 175.

You will get a calculator for this test, right? It might be a little tedious to do this all by hand.

Identity

Member

Registered: 2007-04-18

Posts: 934

Re: Hard problems

Wow thanks... very deep analysis of the problems, I guess making everything into an equation helps a lot. And yeah, calculators allowed. There are a lot of easier questions on the test which hopefully I'll get my marks off... as for these, well... I'll have to think about them

Last edited by Identity (2007-04-28 19:19:16)

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: Hard problems

More direct way of thinking

Last edited by Stanley_Marsh (2007-04-28 20:04:15)

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Numbers are the essence of the Universe

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: Hard problems

Last edited by Stanley_Marsh (2007-04-29 02:27:45)

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Numbers are the essence of the Universe